I need to check if the recipient username is in file /etc/passwd which contains all the users in my class, but I have tried a few different combinations of if statements and grep without success. The best I could come up with is below, but I don't think it's working properly.
My logic behind it is that if the grep is null, the user is invalid.
send_email() { message= address= attachment= validuser=1 until [ "$validuser" = "0" ] do echo "Enter the email address: " read address if [ -z grep $address /etc/passwd ] then validuser=0 else validuser=1 fi echo -n "Enter the subject of the message: " read message echo "" echo "Enter the file you want to attach: " read attachment mail -s "$message" "$address"<"$attachment" done press_enter } 6 Answers
Just do a simple if like this:
if grep -q $address /etc/passwd then echo "OK"; else echo "NOT OK"; fi The -q option is used here just to make grep quiet (don't output...)
4Use getent and check for grep's exit code. Avoid using /etc/passwd. Equivalent in the shell:
getent passwd | grep -q valid_user echo $? Output:
0 And:
getent passwd | grep -q invalid_user echo $? Output:
1 Your piece of code:
if [ -z grep $address /etc/passwd ] You haven't saved the results of grep $address /etc/passwd in a variable. Before putting it in the if statement and then testing the variable to see if it is empty.
You can try it like this:
check_address=`grep $address /etc/passwd` if [ -z "$check_address" ] then validuser=0 else validuser=1 fi The easiest one will be this:
cat test123 # Output: 12345678 cat test123 | grep 123 >/dev/null && echo "grep result exist" || echo "grep result doesn't exist" # Output: grep result exist cat test123 | grep 999 >/dev/null && echo "grep result exist" || echo "grep result doesn't exist" # Output: grep result doesn't exist The -z check is for variable strings, which your grep isn't giving. To give a value from your grep command, enclose it in $():
if [ -z $(grep $address /etc/passwd) ] My problem was that the file I was trying to grep was a binary file. On windows, the first two characters in the file were little squares. On mac, the first two characters were question marks. When I used more or less on the file, I could see it was binary and when I used diff, it responded that the "Binary files foo.log and requirements.txt differ".
I used cat to display the contents of the file, highlighted and copied the text (minus the two question marks at the top, deleted the file, created a new file with touch and then used vi to paste the text back into the new file.
After that, grep worked fine.
